2.3 KiB
Problem: 3-2
Question
2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)
The decomposition of \mathrm{N}_2 \mathrm{O}_5(g) is represented by the equation above. A sample of \mathrm{N}_2 \mathrm{O}_5(g) is monitored as it decomposes, and the concentration of \mathrm{N}_2 \mathrm{O}_5 as a function of time is recorded. The results are shown in the table below.
| Time (s) | \mathrm{[N_2O_5]} |
|---|---|
| 0 | 1.000 |
| 25.0 | 0.801 |
| 50.0 | 0.642 |
| 75.0 | 0.515 |
Calculate the average rate of the reaction between 50.0 and 75.0 seconds.
Solution
a \mathrm{~A}+b \mathrm{~B} \rightarrow c \mathrm{C}+d \mathrm{D}
the rate of reaction is defined as
\color{cyan}\text { rate }=-\frac{1}{a} \frac{\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{b} \frac{\Delta[\mathrm{B}]}{\Delta t}=+\frac{1}{c} \frac{\Delta[\mathrm{C}]}{\Delta t}=+\frac{1}{d} \frac{\Delta[\mathrm{D}]}{\Delta t}
Notice that the rate of change in concentration of each species is divided by its coefficient from the balanced chemical equation ( a, b, c, or d ). This ensures that the calculated reaction rate is the same no matter which reactant or product is monitored for changes in concentration.
In this case, the monitored species was \mathrm{N}_2 \mathrm{O}_5. With that in mind, let's write the reaction rate in terms of the rate of change in concentration of \mathrm{N}_2 \mathrm{O}_5 :
\text { rate }=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta t}
Since the coefficient for \mathrm{N}_2 \mathrm{O}_5 in the balanced equation is 2 , we divided the rate of change in concentration of \mathrm{N}_2 \mathrm{O}_5 by 2 .
Additionally, since \mathrm{N}_2 \mathrm{O}_5 is being consumed in the reaction, we included a negative sign in front of the expression.
Now, let's plug in the information from the table to calculate the average reaction rate between 50.0 and 75.0 seconds:
\begin{aligned}
\text { rate } & =-\frac{1}{2} \frac{(0.515 M-0.642 M)}{(75.0 \mathrm{~s}-50.0 \mathrm{~s})} \\
& =2.54 \times 10^{-3} M \mathrm{~s}^{-1}
\end{aligned}
So, the average rate of the reaction between 50.0 and 75.0 seconds is \bold{2.54 \times 10^{-3} \mathrm{M} \mathrm{s}^{-1}}.